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CIE Jun 2015 v3 Paper 4 Q4
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![(a) Silver sulfate, Ag2SO4, is sparingly soluble in water. The concentration of its saturated solution is 2.5102 mol dm3 at 298 K. (i) Write an expression for the solubility product, Ksp, of Ag2SO4, and state its units. Ksp = units:[1] (ii) Calculate the value for Ksp(Ag2SO4) at 298 K. Ksp =[1] (b) Using Ag2SO4 as an example, complete the following Hess' Law energy cycle relating thelattice energy, enthalpy change of solution, enthalpy change of hydration, , and On your diagram:include the relevant species in the two empty boxes, label each enthalpy change with its appropriate symbol, complete the remaining two arrows showing the correct direction of enthalpy change.Ag2SO4(s)[4]UCLES 2015 9701/43/M/J/15<br />
(c) An electrochemical cell is set up as follows. Pt Fe2(SO4)3(aq) FeSO4(aq) Ag Ag2SO4(aq) Ag2SO4(s) (i) Use the Data Booklet to calculate the value of which electrode is the positive one. under standard conditions, stating =positive electrode:[1] (ii) How would the actual Ecell of the above cell compare to the Explain your answer. under standard conditions?[1] (iii) How would the Ecell of the above cell change, if at all, if a few cm3 of concentrated Na2SO4(aq) were added to the beaker containing Fe3+(aq) + Fe2+(aq),the beaker containing Ag2SO4(aq)?[2] (iv) Explain any changes in Ecell you have stated in (iii).[1] (d) Solutions of iron(III) sulfate are acidic due to the following equilibrium. [Fe(H2O)6]3+(aq) [Fe(H2O)5(OH)]2+(aq) + H+(aq) Ka = 8.9104 mol dm3 Calculate the pH of a 0.1 mol dm3 solution of iron(III) sulfate, Fe2(SO4)3.UCLES 2015 9701/43/M/J/15 pH =[2] [Total: 13] [Turn over<br />](https://learnah.org/wp-content/uploads/2018/08/June_2015_v3_QP_-_Paper_4_CIE_Chemistry_A-level_4-0.jpg)
![Cambridge International A LevelMay/June 2015 Mark Scheme Syllabus Paper 9701 43 Page 3 (a) (i) 1 1 1 [Total: 7] 1 1 1 1 isotope relative abundance 24 Mg 25 Mg 26 Mg 7879 10 1211 (total must add up to 100 %) (ii) e.g. 0.78x24 + 0.10x25 + 0.12x26 = 24.34 (b) (i) nitrates become more stable (down the group) as the ionic radius increases or charge density on cation / ion decreases decreasing its ability to distort / polarise the NO3 / nitrate ion (ii) 4LiNO32Li2O + 4NO2 + O2 (iii) the charge density of the other cations are too small (to polarise the anion sufficiently so the anion is more stable) (a) (i) Ksp = [Ag 2 + (aq)] [SO4 3 (aq)] and units: mol dm 2 x (0.025) = 6.25 x 105 (ii) Ksp = (2 x 0.025) (b) Ag2SO4(s) latt sol 2- hyd or 2- (c) (i) Eo cell (= 0.800.77 =) (+)0.03V and Ag + / Ag or Ag / silver or right (ii) Ecell would be less positive / more negative + because the [Ag (aq)] (in the Ag electrode) is less than 1.0 mol dm (iii)no changeCambridge International Examinations 2015<br />
Page 4 Cambridge International A LevelMay/June 2015 Mark Scheme Syllabus Paper 9701 43more negative / less positive (iv) + (aq)] will decrease the [Ag Eelectrode becomes less positive or due to the common ion effect 3+ (aq)] = 0.2 mol dm [Fe + [H + ]) = 1.9 (or 1.871.89) pH = log([H ] = (c.Ka) = (0.2 x 8.9 x 10 4 ) or 1.33 x 10 2 (mol dm 3 ) (d) (a) protons electrons neutrons 14 (b) (c) (d) no reaction CCl4: GeCl4 and SnCl4: for each steamy fumes evolved or white solid produced GeCl4 + 2H2OGeO2 + 4HCl SnCl4 + 2H2OSnO2 + 4HCl Ge / Sn use dorbitals or Ge / Sn have low lying d orbitals or carbon cannot expand its octet or carbon cannot accommodate more than 4 bonded pairs = +1.69 V and Cl2 / Cl = + 1.36 V is a good reducing agent due to its smaller E value than Cl2 ora is a stronger oxidising agent than Cl2 so Pb with Cl2 reaction is not 2+ is less positive / more negative is a weaker oxidising agent than Cl2 so Sn 2+ with Cl2 reaction is 4+ 2+ 4+ 2+ 2+ 4+ 2+ / Pb / Sn = +0.15V and Pb is oxidised by Cl2 because its Eo Sn Sn or Sn or Pb feasible or Sn feasible SnCl2 + Cl2SnCl4 + Cl2Sn or Sn or SnCl2 + Cl2 + 2H2OSnO2 + 4HCl + 2Cl 4+ 2+ 4+ (e) (i) F = Le (ii) moles of O2(g) = 130 / 24 000 = 5.417 x 10 3 mol moles of electrons needed = 4 x 5.417 x 10 no. of coulombs passed = 1.2 x 30 x 60 or 2160 C no. of electrons passed = 2160 / 1.6 x 10 no. of electrons per mole = 1.35 x 10 / 2.17 x 10 19 22 or 1.35 x 10 22 3 or 2.17 x 10 2 mol 2 = 6.2 x 1023 (mol 1 1 [Total: 13] 1 1 1 1 1 1 1 1 1 1 1 1 1Cambridge International Examinations 2015 [Total: 15]<br />](https://learnah.org/wp-content/uploads/2018/08/June_2015_v3_MS_-_Paper_4_CIE_Chemistry_A-level_4-0.jpg)
